I need to graph: (x^2)+3xy+(y^2)=1 but, using the cot(2theta)=(a-c)/b I got 0, and using cot = 0degrees or cot = 90degrees does not remove the xy term. please help!!!
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Originally Posted by mathmana I need to graph: (x^2)+3xy+(y^2)=1 but, using the cot(2theta)=(a-c)/b I got 0, and using cot = 0degrees or cot = 90degrees does not remove the xy term. please help!!! $\displaystyle \cot (2 \theta) = 0 \Rightarrow 2 \theta = 90^0 \Rightarrow \theta = 45^0$.
Yes, I figured that out yesterday. For some reason I thought cot was y/x. Thanks for responding anyway!!
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